Three Visual Puzzles











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I'm designing a puzzle game, and am trying to play-test some concepts. Can you solve these?



108153767833 = A + B + C.
What are the components and why?



A = ?



B = ?



C = ?



Note for C: As a hint to correct an oversight, the product of the digits in C is 2488320.










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  • 2




    Yep. They are related.
    – NigelMNZ
    Dec 7 at 20:41






  • 1




    yeah never mind, I should have looked closer:-)
    – deep thought
    Dec 7 at 20:42















up vote
7
down vote

favorite
2












I'm designing a puzzle game, and am trying to play-test some concepts. Can you solve these?



108153767833 = A + B + C.
What are the components and why?



A = ?



B = ?



C = ?



Note for C: As a hint to correct an oversight, the product of the digits in C is 2488320.










share|improve this question




















  • 2




    Yep. They are related.
    – NigelMNZ
    Dec 7 at 20:41






  • 1




    yeah never mind, I should have looked closer:-)
    – deep thought
    Dec 7 at 20:42













up vote
7
down vote

favorite
2









up vote
7
down vote

favorite
2






2





I'm designing a puzzle game, and am trying to play-test some concepts. Can you solve these?



108153767833 = A + B + C.
What are the components and why?



A = ?



B = ?



C = ?



Note for C: As a hint to correct an oversight, the product of the digits in C is 2488320.










share|improve this question















I'm designing a puzzle game, and am trying to play-test some concepts. Can you solve these?



108153767833 = A + B + C.
What are the components and why?



A = ?



B = ?



C = ?



Note for C: As a hint to correct an oversight, the product of the digits in C is 2488320.







mathematics visual mazes alphametic






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 8 at 6:19

























asked Dec 7 at 20:38









NigelMNZ

985




985








  • 2




    Yep. They are related.
    – NigelMNZ
    Dec 7 at 20:41






  • 1




    yeah never mind, I should have looked closer:-)
    – deep thought
    Dec 7 at 20:42














  • 2




    Yep. They are related.
    – NigelMNZ
    Dec 7 at 20:41






  • 1




    yeah never mind, I should have looked closer:-)
    – deep thought
    Dec 7 at 20:42








2




2




Yep. They are related.
– NigelMNZ
Dec 7 at 20:41




Yep. They are related.
– NigelMNZ
Dec 7 at 20:41




1




1




yeah never mind, I should have looked closer:-)
– deep thought
Dec 7 at 20:42




yeah never mind, I should have looked closer:-)
– deep thought
Dec 7 at 20:42










3 Answers
3






active

oldest

votes

















up vote
6
down vote



accepted










Partial:



C




Mazes with the following rules:

1. Take the shortest path possible starting at green and ending at red.

2. If yellow dots, all must be "retrieved" before reaching red.

3. Red cannot be crossed, thus only reached once.

4. Colored squares are "locked doors" that are only passable once you have "retrieved" the associated colored triangle.

Based on these rules and clarification provided by a hint and a comment, we can now solve the last maze, which gives us a solution of
$99453888111$






A




Found by @deep_thought. Please review their answer.






B




After working through this and making a mess of a wall of text, I found that @JonMark_Perry was better able to succinctly describe the process.

Let's look at #1 for an example though:

Example 1. a_1_z -> pp=11

p falls in the range of a to z, thus p is 1. As p is printed twice, the result is 11.


Example 2. a_1_ij_2_op_3_z -> pi = 31 ty =33

set p-z includes p, t, and y, thus p, t, and y are all 3.

set a-i includes i, thus i is 1


Example 3 gets a little interesting.
Here, q_1_k translates to ^k-q = 1, or letters not in the set of k-q get +1.

3 also introduces the idea that set values can stack.

As such, p is in set a-p and not in set k-q, thus is 1 + 1 or 2.

l is in set a-p, k-q, l-z, thus is 1 + -1 or 2. (Doesn't get the 1 from ^k-q as l is not not in k-q.


Thus brings us to Example 5, which we must solve to find B.
The rule sets are as follows:

a-z = 1

a-m = 1 n-z = 2

a-p = 2 q-z = 1

a-e = 1 f-z = 3

a-j = 1 k-s = 2 t-z = 3

a-p = -2 q-z = -1

These thus result in the following letter values:

p = 1 2 2 3 2 -2 = 8

i = 3 1 = 4

t = 1 2 3 3 = 9

y = 1 2 3 3 = 9

l = 1 1 3 2 = 7

e = 1 1 1 1 = 4

a = 1 1 1 1 = 4

s = 1 2 3 2 = 8

And when the word pityplease is decoded, we get

B = $8499874484$

However, this value appears to be incorrect.

Actually, what if a_-2_pq_-1_z were instead just a_-1_z:

p = 1 2 2 3 2 -1 = 9

i = 1 1 2 3 1 -1 = 7

t = 1 2 1 3 3 -1 = 9

y = 1 2 1 3 3 -1 = 9

l = 1 1 2 3 2 -1 = 8

e = 1 1 2 1 1 -1 = 5

a = 1 1 2 1 1 -1 = 5

s = 1 2 1 3 2 -1 = 8
$9799985585$

Perfect! Just what we needed!




Answer:




If T is $109253878934$

And A is confirmed $5238$

And C is confirmed $99453888111$

Then B must be $109253878934$ - $5238$ - $99453888111$ = $9799985585$

However, I can't get my B formula to match.

Actually, if we use the slightly modified rule for B5, then we get the desired result!

Technically not correct, but I'm happy with it for now.







share|improve this answer























  • Looks like you are correct. Totally an oversight on my part. Nice job finding that. My intended solution was none of those three, interestingly enough.
    – NigelMNZ
    Dec 7 at 21:30






  • 1




    you can have 94492888111 as well.
    – JonMark Perry
    Dec 7 at 21:32










  • For clarity, my previous comment was in reference to your solution for C.
    – NigelMNZ
    Dec 7 at 21:58












  • Getting somewhere, but not quite. Check out my added note if you'd like to find the exact answer for C.
    – NigelMNZ
    Dec 7 at 22:50










  • Yes but my current C value actually works for that hint @NigelMNZ.
    – Dorrulf
    Dec 7 at 22:51




















up vote
5
down vote













Partial answer



For A,




if we ignore tilt and position, then I can fit the four numbers at the top with: colour is power of ten, shape is digit (star is one), filled is positive, empty is negative; sum all shapes.

- filled red star = 1

- filled red triangle and filled red square = 3 + 4 = 7

- filled blue pentagon and empty red heptagon = 50 - 7 = 43

- filled green triangle and empty blue square and empty red enneagon = 300 - 40 - 9 = 251

This gives P^1 = 157, P^2 = 857, P^3 = 1201. These are prime factors of the number at the bottom. P^4 is the missing factor is 3023. Their sum is A = 5238.




Part B: no clue



Part C: already solved by @Dorrulf






share|improve this answer



















  • 1




    Nicely done for A!
    – Dorrulf
    Dec 7 at 22:05






  • 1




    Indeed, very nice.
    – NigelMNZ
    Dec 7 at 22:16


















up vote
5
down vote













B.




Every a_b_c triple forms a range in the alphabet that scores b points, for example a_1_m means that every letter in [a,m] scores 1 point, with wrapround if the range is backwards. I get $B=8699874484$.







share|improve this answer





















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    3 Answers
    3






    active

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    3 Answers
    3






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes








    up vote
    6
    down vote



    accepted










    Partial:



    C




    Mazes with the following rules:

    1. Take the shortest path possible starting at green and ending at red.

    2. If yellow dots, all must be "retrieved" before reaching red.

    3. Red cannot be crossed, thus only reached once.

    4. Colored squares are "locked doors" that are only passable once you have "retrieved" the associated colored triangle.

    Based on these rules and clarification provided by a hint and a comment, we can now solve the last maze, which gives us a solution of
    $99453888111$






    A




    Found by @deep_thought. Please review their answer.






    B




    After working through this and making a mess of a wall of text, I found that @JonMark_Perry was better able to succinctly describe the process.

    Let's look at #1 for an example though:

    Example 1. a_1_z -> pp=11

    p falls in the range of a to z, thus p is 1. As p is printed twice, the result is 11.


    Example 2. a_1_ij_2_op_3_z -> pi = 31 ty =33

    set p-z includes p, t, and y, thus p, t, and y are all 3.

    set a-i includes i, thus i is 1


    Example 3 gets a little interesting.
    Here, q_1_k translates to ^k-q = 1, or letters not in the set of k-q get +1.

    3 also introduces the idea that set values can stack.

    As such, p is in set a-p and not in set k-q, thus is 1 + 1 or 2.

    l is in set a-p, k-q, l-z, thus is 1 + -1 or 2. (Doesn't get the 1 from ^k-q as l is not not in k-q.


    Thus brings us to Example 5, which we must solve to find B.
    The rule sets are as follows:

    a-z = 1

    a-m = 1 n-z = 2

    a-p = 2 q-z = 1

    a-e = 1 f-z = 3

    a-j = 1 k-s = 2 t-z = 3

    a-p = -2 q-z = -1

    These thus result in the following letter values:

    p = 1 2 2 3 2 -2 = 8

    i = 3 1 = 4

    t = 1 2 3 3 = 9

    y = 1 2 3 3 = 9

    l = 1 1 3 2 = 7

    e = 1 1 1 1 = 4

    a = 1 1 1 1 = 4

    s = 1 2 3 2 = 8

    And when the word pityplease is decoded, we get

    B = $8499874484$

    However, this value appears to be incorrect.

    Actually, what if a_-2_pq_-1_z were instead just a_-1_z:

    p = 1 2 2 3 2 -1 = 9

    i = 1 1 2 3 1 -1 = 7

    t = 1 2 1 3 3 -1 = 9

    y = 1 2 1 3 3 -1 = 9

    l = 1 1 2 3 2 -1 = 8

    e = 1 1 2 1 1 -1 = 5

    a = 1 1 2 1 1 -1 = 5

    s = 1 2 1 3 2 -1 = 8
    $9799985585$

    Perfect! Just what we needed!




    Answer:




    If T is $109253878934$

    And A is confirmed $5238$

    And C is confirmed $99453888111$

    Then B must be $109253878934$ - $5238$ - $99453888111$ = $9799985585$

    However, I can't get my B formula to match.

    Actually, if we use the slightly modified rule for B5, then we get the desired result!

    Technically not correct, but I'm happy with it for now.







    share|improve this answer























    • Looks like you are correct. Totally an oversight on my part. Nice job finding that. My intended solution was none of those three, interestingly enough.
      – NigelMNZ
      Dec 7 at 21:30






    • 1




      you can have 94492888111 as well.
      – JonMark Perry
      Dec 7 at 21:32










    • For clarity, my previous comment was in reference to your solution for C.
      – NigelMNZ
      Dec 7 at 21:58












    • Getting somewhere, but not quite. Check out my added note if you'd like to find the exact answer for C.
      – NigelMNZ
      Dec 7 at 22:50










    • Yes but my current C value actually works for that hint @NigelMNZ.
      – Dorrulf
      Dec 7 at 22:51

















    up vote
    6
    down vote



    accepted










    Partial:



    C




    Mazes with the following rules:

    1. Take the shortest path possible starting at green and ending at red.

    2. If yellow dots, all must be "retrieved" before reaching red.

    3. Red cannot be crossed, thus only reached once.

    4. Colored squares are "locked doors" that are only passable once you have "retrieved" the associated colored triangle.

    Based on these rules and clarification provided by a hint and a comment, we can now solve the last maze, which gives us a solution of
    $99453888111$






    A




    Found by @deep_thought. Please review their answer.






    B




    After working through this and making a mess of a wall of text, I found that @JonMark_Perry was better able to succinctly describe the process.

    Let's look at #1 for an example though:

    Example 1. a_1_z -> pp=11

    p falls in the range of a to z, thus p is 1. As p is printed twice, the result is 11.


    Example 2. a_1_ij_2_op_3_z -> pi = 31 ty =33

    set p-z includes p, t, and y, thus p, t, and y are all 3.

    set a-i includes i, thus i is 1


    Example 3 gets a little interesting.
    Here, q_1_k translates to ^k-q = 1, or letters not in the set of k-q get +1.

    3 also introduces the idea that set values can stack.

    As such, p is in set a-p and not in set k-q, thus is 1 + 1 or 2.

    l is in set a-p, k-q, l-z, thus is 1 + -1 or 2. (Doesn't get the 1 from ^k-q as l is not not in k-q.


    Thus brings us to Example 5, which we must solve to find B.
    The rule sets are as follows:

    a-z = 1

    a-m = 1 n-z = 2

    a-p = 2 q-z = 1

    a-e = 1 f-z = 3

    a-j = 1 k-s = 2 t-z = 3

    a-p = -2 q-z = -1

    These thus result in the following letter values:

    p = 1 2 2 3 2 -2 = 8

    i = 3 1 = 4

    t = 1 2 3 3 = 9

    y = 1 2 3 3 = 9

    l = 1 1 3 2 = 7

    e = 1 1 1 1 = 4

    a = 1 1 1 1 = 4

    s = 1 2 3 2 = 8

    And when the word pityplease is decoded, we get

    B = $8499874484$

    However, this value appears to be incorrect.

    Actually, what if a_-2_pq_-1_z were instead just a_-1_z:

    p = 1 2 2 3 2 -1 = 9

    i = 1 1 2 3 1 -1 = 7

    t = 1 2 1 3 3 -1 = 9

    y = 1 2 1 3 3 -1 = 9

    l = 1 1 2 3 2 -1 = 8

    e = 1 1 2 1 1 -1 = 5

    a = 1 1 2 1 1 -1 = 5

    s = 1 2 1 3 2 -1 = 8
    $9799985585$

    Perfect! Just what we needed!




    Answer:




    If T is $109253878934$

    And A is confirmed $5238$

    And C is confirmed $99453888111$

    Then B must be $109253878934$ - $5238$ - $99453888111$ = $9799985585$

    However, I can't get my B formula to match.

    Actually, if we use the slightly modified rule for B5, then we get the desired result!

    Technically not correct, but I'm happy with it for now.







    share|improve this answer























    • Looks like you are correct. Totally an oversight on my part. Nice job finding that. My intended solution was none of those three, interestingly enough.
      – NigelMNZ
      Dec 7 at 21:30






    • 1




      you can have 94492888111 as well.
      – JonMark Perry
      Dec 7 at 21:32










    • For clarity, my previous comment was in reference to your solution for C.
      – NigelMNZ
      Dec 7 at 21:58












    • Getting somewhere, but not quite. Check out my added note if you'd like to find the exact answer for C.
      – NigelMNZ
      Dec 7 at 22:50










    • Yes but my current C value actually works for that hint @NigelMNZ.
      – Dorrulf
      Dec 7 at 22:51















    up vote
    6
    down vote



    accepted







    up vote
    6
    down vote



    accepted






    Partial:



    C




    Mazes with the following rules:

    1. Take the shortest path possible starting at green and ending at red.

    2. If yellow dots, all must be "retrieved" before reaching red.

    3. Red cannot be crossed, thus only reached once.

    4. Colored squares are "locked doors" that are only passable once you have "retrieved" the associated colored triangle.

    Based on these rules and clarification provided by a hint and a comment, we can now solve the last maze, which gives us a solution of
    $99453888111$






    A




    Found by @deep_thought. Please review their answer.






    B




    After working through this and making a mess of a wall of text, I found that @JonMark_Perry was better able to succinctly describe the process.

    Let's look at #1 for an example though:

    Example 1. a_1_z -> pp=11

    p falls in the range of a to z, thus p is 1. As p is printed twice, the result is 11.


    Example 2. a_1_ij_2_op_3_z -> pi = 31 ty =33

    set p-z includes p, t, and y, thus p, t, and y are all 3.

    set a-i includes i, thus i is 1


    Example 3 gets a little interesting.
    Here, q_1_k translates to ^k-q = 1, or letters not in the set of k-q get +1.

    3 also introduces the idea that set values can stack.

    As such, p is in set a-p and not in set k-q, thus is 1 + 1 or 2.

    l is in set a-p, k-q, l-z, thus is 1 + -1 or 2. (Doesn't get the 1 from ^k-q as l is not not in k-q.


    Thus brings us to Example 5, which we must solve to find B.
    The rule sets are as follows:

    a-z = 1

    a-m = 1 n-z = 2

    a-p = 2 q-z = 1

    a-e = 1 f-z = 3

    a-j = 1 k-s = 2 t-z = 3

    a-p = -2 q-z = -1

    These thus result in the following letter values:

    p = 1 2 2 3 2 -2 = 8

    i = 3 1 = 4

    t = 1 2 3 3 = 9

    y = 1 2 3 3 = 9

    l = 1 1 3 2 = 7

    e = 1 1 1 1 = 4

    a = 1 1 1 1 = 4

    s = 1 2 3 2 = 8

    And when the word pityplease is decoded, we get

    B = $8499874484$

    However, this value appears to be incorrect.

    Actually, what if a_-2_pq_-1_z were instead just a_-1_z:

    p = 1 2 2 3 2 -1 = 9

    i = 1 1 2 3 1 -1 = 7

    t = 1 2 1 3 3 -1 = 9

    y = 1 2 1 3 3 -1 = 9

    l = 1 1 2 3 2 -1 = 8

    e = 1 1 2 1 1 -1 = 5

    a = 1 1 2 1 1 -1 = 5

    s = 1 2 1 3 2 -1 = 8
    $9799985585$

    Perfect! Just what we needed!




    Answer:




    If T is $109253878934$

    And A is confirmed $5238$

    And C is confirmed $99453888111$

    Then B must be $109253878934$ - $5238$ - $99453888111$ = $9799985585$

    However, I can't get my B formula to match.

    Actually, if we use the slightly modified rule for B5, then we get the desired result!

    Technically not correct, but I'm happy with it for now.







    share|improve this answer














    Partial:



    C




    Mazes with the following rules:

    1. Take the shortest path possible starting at green and ending at red.

    2. If yellow dots, all must be "retrieved" before reaching red.

    3. Red cannot be crossed, thus only reached once.

    4. Colored squares are "locked doors" that are only passable once you have "retrieved" the associated colored triangle.

    Based on these rules and clarification provided by a hint and a comment, we can now solve the last maze, which gives us a solution of
    $99453888111$






    A




    Found by @deep_thought. Please review their answer.






    B




    After working through this and making a mess of a wall of text, I found that @JonMark_Perry was better able to succinctly describe the process.

    Let's look at #1 for an example though:

    Example 1. a_1_z -> pp=11

    p falls in the range of a to z, thus p is 1. As p is printed twice, the result is 11.


    Example 2. a_1_ij_2_op_3_z -> pi = 31 ty =33

    set p-z includes p, t, and y, thus p, t, and y are all 3.

    set a-i includes i, thus i is 1


    Example 3 gets a little interesting.
    Here, q_1_k translates to ^k-q = 1, or letters not in the set of k-q get +1.

    3 also introduces the idea that set values can stack.

    As such, p is in set a-p and not in set k-q, thus is 1 + 1 or 2.

    l is in set a-p, k-q, l-z, thus is 1 + -1 or 2. (Doesn't get the 1 from ^k-q as l is not not in k-q.


    Thus brings us to Example 5, which we must solve to find B.
    The rule sets are as follows:

    a-z = 1

    a-m = 1 n-z = 2

    a-p = 2 q-z = 1

    a-e = 1 f-z = 3

    a-j = 1 k-s = 2 t-z = 3

    a-p = -2 q-z = -1

    These thus result in the following letter values:

    p = 1 2 2 3 2 -2 = 8

    i = 3 1 = 4

    t = 1 2 3 3 = 9

    y = 1 2 3 3 = 9

    l = 1 1 3 2 = 7

    e = 1 1 1 1 = 4

    a = 1 1 1 1 = 4

    s = 1 2 3 2 = 8

    And when the word pityplease is decoded, we get

    B = $8499874484$

    However, this value appears to be incorrect.

    Actually, what if a_-2_pq_-1_z were instead just a_-1_z:

    p = 1 2 2 3 2 -1 = 9

    i = 1 1 2 3 1 -1 = 7

    t = 1 2 1 3 3 -1 = 9

    y = 1 2 1 3 3 -1 = 9

    l = 1 1 2 3 2 -1 = 8

    e = 1 1 2 1 1 -1 = 5

    a = 1 1 2 1 1 -1 = 5

    s = 1 2 1 3 2 -1 = 8
    $9799985585$

    Perfect! Just what we needed!




    Answer:




    If T is $109253878934$

    And A is confirmed $5238$

    And C is confirmed $99453888111$

    Then B must be $109253878934$ - $5238$ - $99453888111$ = $9799985585$

    However, I can't get my B formula to match.

    Actually, if we use the slightly modified rule for B5, then we get the desired result!

    Technically not correct, but I'm happy with it for now.








    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Dec 7 at 23:56

























    answered Dec 7 at 21:12









    Dorrulf

    2,07718




    2,07718












    • Looks like you are correct. Totally an oversight on my part. Nice job finding that. My intended solution was none of those three, interestingly enough.
      – NigelMNZ
      Dec 7 at 21:30






    • 1




      you can have 94492888111 as well.
      – JonMark Perry
      Dec 7 at 21:32










    • For clarity, my previous comment was in reference to your solution for C.
      – NigelMNZ
      Dec 7 at 21:58












    • Getting somewhere, but not quite. Check out my added note if you'd like to find the exact answer for C.
      – NigelMNZ
      Dec 7 at 22:50










    • Yes but my current C value actually works for that hint @NigelMNZ.
      – Dorrulf
      Dec 7 at 22:51




















    • Looks like you are correct. Totally an oversight on my part. Nice job finding that. My intended solution was none of those three, interestingly enough.
      – NigelMNZ
      Dec 7 at 21:30






    • 1




      you can have 94492888111 as well.
      – JonMark Perry
      Dec 7 at 21:32










    • For clarity, my previous comment was in reference to your solution for C.
      – NigelMNZ
      Dec 7 at 21:58












    • Getting somewhere, but not quite. Check out my added note if you'd like to find the exact answer for C.
      – NigelMNZ
      Dec 7 at 22:50










    • Yes but my current C value actually works for that hint @NigelMNZ.
      – Dorrulf
      Dec 7 at 22:51


















    Looks like you are correct. Totally an oversight on my part. Nice job finding that. My intended solution was none of those three, interestingly enough.
    – NigelMNZ
    Dec 7 at 21:30




    Looks like you are correct. Totally an oversight on my part. Nice job finding that. My intended solution was none of those three, interestingly enough.
    – NigelMNZ
    Dec 7 at 21:30




    1




    1




    you can have 94492888111 as well.
    – JonMark Perry
    Dec 7 at 21:32




    you can have 94492888111 as well.
    – JonMark Perry
    Dec 7 at 21:32












    For clarity, my previous comment was in reference to your solution for C.
    – NigelMNZ
    Dec 7 at 21:58






    For clarity, my previous comment was in reference to your solution for C.
    – NigelMNZ
    Dec 7 at 21:58














    Getting somewhere, but not quite. Check out my added note if you'd like to find the exact answer for C.
    – NigelMNZ
    Dec 7 at 22:50




    Getting somewhere, but not quite. Check out my added note if you'd like to find the exact answer for C.
    – NigelMNZ
    Dec 7 at 22:50












    Yes but my current C value actually works for that hint @NigelMNZ.
    – Dorrulf
    Dec 7 at 22:51






    Yes but my current C value actually works for that hint @NigelMNZ.
    – Dorrulf
    Dec 7 at 22:51












    up vote
    5
    down vote













    Partial answer



    For A,




    if we ignore tilt and position, then I can fit the four numbers at the top with: colour is power of ten, shape is digit (star is one), filled is positive, empty is negative; sum all shapes.

    - filled red star = 1

    - filled red triangle and filled red square = 3 + 4 = 7

    - filled blue pentagon and empty red heptagon = 50 - 7 = 43

    - filled green triangle and empty blue square and empty red enneagon = 300 - 40 - 9 = 251

    This gives P^1 = 157, P^2 = 857, P^3 = 1201. These are prime factors of the number at the bottom. P^4 is the missing factor is 3023. Their sum is A = 5238.




    Part B: no clue



    Part C: already solved by @Dorrulf






    share|improve this answer



















    • 1




      Nicely done for A!
      – Dorrulf
      Dec 7 at 22:05






    • 1




      Indeed, very nice.
      – NigelMNZ
      Dec 7 at 22:16















    up vote
    5
    down vote













    Partial answer



    For A,




    if we ignore tilt and position, then I can fit the four numbers at the top with: colour is power of ten, shape is digit (star is one), filled is positive, empty is negative; sum all shapes.

    - filled red star = 1

    - filled red triangle and filled red square = 3 + 4 = 7

    - filled blue pentagon and empty red heptagon = 50 - 7 = 43

    - filled green triangle and empty blue square and empty red enneagon = 300 - 40 - 9 = 251

    This gives P^1 = 157, P^2 = 857, P^3 = 1201. These are prime factors of the number at the bottom. P^4 is the missing factor is 3023. Their sum is A = 5238.




    Part B: no clue



    Part C: already solved by @Dorrulf






    share|improve this answer



















    • 1




      Nicely done for A!
      – Dorrulf
      Dec 7 at 22:05






    • 1




      Indeed, very nice.
      – NigelMNZ
      Dec 7 at 22:16













    up vote
    5
    down vote










    up vote
    5
    down vote









    Partial answer



    For A,




    if we ignore tilt and position, then I can fit the four numbers at the top with: colour is power of ten, shape is digit (star is one), filled is positive, empty is negative; sum all shapes.

    - filled red star = 1

    - filled red triangle and filled red square = 3 + 4 = 7

    - filled blue pentagon and empty red heptagon = 50 - 7 = 43

    - filled green triangle and empty blue square and empty red enneagon = 300 - 40 - 9 = 251

    This gives P^1 = 157, P^2 = 857, P^3 = 1201. These are prime factors of the number at the bottom. P^4 is the missing factor is 3023. Their sum is A = 5238.




    Part B: no clue



    Part C: already solved by @Dorrulf






    share|improve this answer














    Partial answer



    For A,




    if we ignore tilt and position, then I can fit the four numbers at the top with: colour is power of ten, shape is digit (star is one), filled is positive, empty is negative; sum all shapes.

    - filled red star = 1

    - filled red triangle and filled red square = 3 + 4 = 7

    - filled blue pentagon and empty red heptagon = 50 - 7 = 43

    - filled green triangle and empty blue square and empty red enneagon = 300 - 40 - 9 = 251

    This gives P^1 = 157, P^2 = 857, P^3 = 1201. These are prime factors of the number at the bottom. P^4 is the missing factor is 3023. Their sum is A = 5238.




    Part B: no clue



    Part C: already solved by @Dorrulf







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Dec 7 at 22:02

























    answered Dec 7 at 21:52









    deep thought

    2,406632




    2,406632








    • 1




      Nicely done for A!
      – Dorrulf
      Dec 7 at 22:05






    • 1




      Indeed, very nice.
      – NigelMNZ
      Dec 7 at 22:16














    • 1




      Nicely done for A!
      – Dorrulf
      Dec 7 at 22:05






    • 1




      Indeed, very nice.
      – NigelMNZ
      Dec 7 at 22:16








    1




    1




    Nicely done for A!
    – Dorrulf
    Dec 7 at 22:05




    Nicely done for A!
    – Dorrulf
    Dec 7 at 22:05




    1




    1




    Indeed, very nice.
    – NigelMNZ
    Dec 7 at 22:16




    Indeed, very nice.
    – NigelMNZ
    Dec 7 at 22:16










    up vote
    5
    down vote













    B.




    Every a_b_c triple forms a range in the alphabet that scores b points, for example a_1_m means that every letter in [a,m] scores 1 point, with wrapround if the range is backwards. I get $B=8699874484$.







    share|improve this answer

























      up vote
      5
      down vote













      B.




      Every a_b_c triple forms a range in the alphabet that scores b points, for example a_1_m means that every letter in [a,m] scores 1 point, with wrapround if the range is backwards. I get $B=8699874484$.







      share|improve this answer























        up vote
        5
        down vote










        up vote
        5
        down vote









        B.




        Every a_b_c triple forms a range in the alphabet that scores b points, for example a_1_m means that every letter in [a,m] scores 1 point, with wrapround if the range is backwards. I get $B=8699874484$.







        share|improve this answer












        B.




        Every a_b_c triple forms a range in the alphabet that scores b points, for example a_1_m means that every letter in [a,m] scores 1 point, with wrapround if the range is backwards. I get $B=8699874484$.








        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Dec 7 at 22:34









        JonMark Perry

        17.1k63281




        17.1k63281






























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